Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate a limit of exponential function Calculate this limit:
$$
\lim_{x \to \infty } = \left(\frac{1}{5} + \frac{1}{5x}\right)^{\frac{x}{5}}
$$
I did this:
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left[\left(1+\frac{1}{x}\right)^{x}\right]^\frac{1}{5}
$$
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\frac{5}{5}\ri... | Write
$$
\left(\frac{1+x}{5x}\right)^{x/5} = e^{\frac{x}{5}\ln\left(\frac{1+x}{5x}\right)} = e^{-\frac{x}{5}\ln\left(\frac{5x}{1+x}\right)} = e^{-\frac{x}{5}\ln\left(5-\frac{5}{1+x}\right)}
$$
Observing that $\ln\left(5-\frac{5}{1+x}\right)\xrightarrow[x\to\infty]{}\ln 5$, can you conclude by composing limits?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ Evaluation of $\displaystyle \int \frac{\sqrt{1+x^4}}{1-x^4}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{\sqrt{1+x^4}}{1-x^4}dx\;,$ Then We can write the above Integral as $$\displaystyle \int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}dx = \frac{1... | We can write the integral as
$$
I=\int\frac{\sqrt{1+x^4}}{1-x^4}\textrm{d}x=\int\frac{\left(1+x^4\right)}{\left(1-x^4\right)\sqrt{1+x^4}}\textrm{d}x.
$$
Let $t=\frac{1+x^4}{1-x^4}$ so that $x^4=\frac{t+1}{t-1}$ and $\textrm{d}x=\pm\left(\frac{t+1}{t-1}\right)^{\frac{1}{4}}\cdot\frac{-1}{2(t-1)(t+1)}\textrm{d}t$.
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Transforming linear combination of the cosine and sine function In the proof of Transforming $a\cos\left(\, x\,\right)+b\sin\left(\, x\right)$ to $r\cos\left(\,\phi - x\,\right)$
\begin{align}
a\cos\left(\, x\,\right) + b\sin\left(\, x\,\right)
&=\,\sqrt{\,a^{2} + b^{2}\,}\,
\left[\,\frac{a}{\,\sqrt{\, a^{2} + b^{2}\,}... | The idea behind the proof comes from noting that
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\dfrac{b}{\sqrt{a^2+b^2}}\right)^2=1.\tag1$$
This is analogous to the Pythagorean identity
$$\cos^2\phi+\sin^2\phi=1.\tag2$$
So for all $a$ and $b$ we can get an expression that satisfies $(2)$ by just factoring out $\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of roots: Vieta's Formula The roots of the equation $x^4-5x^2+2x-1=0$ are $\alpha, \beta, \gamma, \delta$. Let $S_n=\alpha^n +\beta^n+\gamma^n+\delta^n$
Show that $S_{n+4}-5S_{n+2}+2S_{n+1}-S_{n}=0$
I have no idea how to approach this. Could someone point me in the right direction?
| Hint. $$ \begin {align*} S_{n+4} - 5S_{n+2} + 2S_{n+1} - S_n &= \alpha^{n+4} + \beta^{n+4} + \gamma^{n+4} + \delta^{n+4} - 5 \cdot \left( \alpha^{n+2} + \beta^{n+2} + \gamma^{n+2} + \delta^{n+2} \right) + 2 \cdot \left( \alpha^{n+1} + \beta^{n+1} + \gamma^{n+1} + \delta^{n+1} \right) - \left( \alpha^n + \beta^n + \gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Power series for the rational function $(1+x)^3/(1-x)^3$
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$
I tried with the partial frationaising the expression that gives me
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$
how to proceed further on this havi... | As shown in this answer, $\binom{-n\vphantom{1}}{k}=(-1)^k\binom{n+k-1}{k}$, therefore,
$$
\begin{align}
\frac{(1+x)^3}{(1-x)^3}
&=(1+x)^3\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\
&=(1+x)^3\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{k=0}^\infty\sum_{j=0}^3\binom{3}{j}\binom{k-j+2}{k-j}x^k\\
&=1+\sum_{k=1}^\infty(4k^2+2)x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find Sum of $\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$. Prove that it converges. Question : For $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$$
a. Prove it converges
b. Find the sum
My Try
$
= \sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)\\
= \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 -... | Hint
$$\ln\left(1-\frac1{n^2}\right)=\ln\left(\frac{(n-1)(n+1)}{n^2}\right)=\ln\left(\frac{n-1}{n}\right)-\ln\left(\frac{n}{n+1}\right)=u_n-u_{n+1}$$
and then telescope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$ Good morning
help me to show the following inequality
for all $a$, $b$ two positive real numbers
$$a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$$
thanks you
| Let $a=\frac{x^2}{z},\, b=\frac{y^2}{z}$ then $x,y \geqq 0$ and $z>0.$ Then we have$:$
$$z^6 \cdot (\text{LHS}-\text{RHS})={x}^{3}{y}^{3} \left( x-y \right) ^{2} \left( {x}^{3}y+{x}^{2}{y}^{2}+ x{y}^{3}+{z}^{2} \right) + \left( xy+z \right) \left( {x}^{2}{y}^{2}+ xyz+{z}^{2} \right) \left( xy-z \right) ^{2} \left( {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$b_{n}$ is increasing I think there is misunderstanding in my last post because its contain three questions so i will post question by question step by step
An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_... | Notice that $\dfrac{b_{n+1}}{b_n}=\dfrac{\cos^2\left(\dfrac{\pi}{2^{n+1}}\right)}{\cos\left(\dfrac{\pi}{2^n}\right)}=\dfrac{\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2}}{\cos\left(\frac{\pi}{2^n}\right)}=\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2\cos\left(\dfrac{\pi}{2^n}\right)}>\dfrac{2\cos\left(\dfrac{\pi}{2^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve IVP using Laplace transform(of matrix)? $$x' = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3&2 & 1\end{bmatrix} x, ~~ x{(0)} = \begin{bmatrix} 2 \\ -1 \\ 1\end{bmatrix}$$
I am having very hard time solving this question using Laplace transform. I don't know how to do it for $3\times 3$ matrix. Would anyone ... | We are given:
$$X'(t) = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2\\ 3 & 2 & 1\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 2 \\ -1\\ 1\end{bmatrix}$$
We can write this as:
$$\tag 1 \begin{align} x' &= x \\ y' &= 2x+y-2z \\ z' &= 3x +2y + z \end{align}$$
Taking the Laplace tran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that a number is an integer. Prove that the following number is an integer:
$$\left( \dfrac{76}{\dfrac{1}{\sqrt[\large{3}]{77}-\sqrt[\large{3}]{75}}-\sqrt[\large{3}]{5775}}+\dfrac{1}{\dfrac{76}{\sqrt[\large{3}]{77}+\sqrt[\large{3}]{75}}+\sqrt[\large{3}]{5775}}\right)^{\large{3}}$$
How can I prove it?
| we have $$ \left( 76\, \left( \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt [3]{5775} \right) ^{-1}+ \left( 76\, \left( \sqrt [3]{77}+\sqrt
[3]{75} \right) ^{-1}+\sqrt [3]{5775} \right) ^{-1} \right) ^{3}
$$
after expanding we obtain
$$438976\, \left( \left( \sqrt [3]{77}-\sqrt [3]{75} \right) ^{-1}-
\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
Finding the area of a triangle, given the distance between center of incircle and circumscribed circle Consider the following depiction:
$ABC$ is an isosceles triangle ($AB=AC$), where the two angles opposite the equal sides are equal $\beta$ ($\beta>60$), and $AD$ perpendicular to $BC$.
$O$ is the center of the circu... | Assuming $BD=1$, we have $AB=AC=\frac{1}{\cos\beta}$, $[ABC]=\tan\beta$,
$$ R = \frac{AB}{2\sin\beta} = \frac{1}{\sin(2\beta)}$$
and:
$$ r = \frac{2[ABC]}{AB+AC+BC} = \frac{\tan\beta}{1+\frac{1}{\cos\beta}}=\frac{\sin\beta}{1+\cos\beta}=\tan\frac{\beta}{2}. $$
Moreover, by Euler's theorem we have:
$$ k^2 = OM^2 = R^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{... | to shorten the working, since the function is only well-defined for $x \in (2,3)$ we may substitute
$$
3-x \to s
$$
giving
$$
I=\int\sqrt{\frac{x-3}{2-x}} .dx = - \int\sqrt{\frac{s}{1-s}} .ds
$$
now substitute
$$
s \to \sin^2 \theta \\
ds \to 2\sin\theta \cos\theta\cdot d\theta
$$
so
$$
I = -\int2\sin^2 \theta \cdot d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2... | You can also use Taylor series. Start writing $$A=\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}$$ $$\frac Ax=\sqrt{1+\frac{5x-2}{x^2}}-2\sqrt{1-\frac{3x-7}{4x^2}}+\sqrt{1+\frac{7x+5}{x^2}}$$ Now, remember that, when $y$ is small compared to $1$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$.
So replace ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Consecutive cubes equal to a square $\frac{1}{8}ab(a^2+b^2-1) = y^2$, and Pythagorean triples If we wish that the sum of $b$ consecutive cubes with initial cube $c_k=\tfrac{1}{2}(1+a-b)$ is equal to a square, then we have the rather simple equation,
$$F_k=\tfrac{1}{8}ab(a^2+b^2-1) = y^2$$
It seems only three integral f... | As pointed out by Barto, appropriate substitutions can transform the problem into solving a Pell-like, or even a Pell equation. For example, let,
$$a =2bn^2,\quad y = bnu/2$$
and $F_k=\frac{1}{8}ab(a^2+b^2-1) = y^2\,$ transforms into the Pell,
$$u^2-(4n^4+1)b^2 = -1\tag1$$
An initial solution is $u_0 = 2n^2,\; b_0 = 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trace of a matrix $A$ with $A^2=I$ Let $A$ be a complex-value square matrix with $A^2=I$ identity.
Then is the trace of $A$ a real value?
| $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\implies A^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies$$
$$\begin{align}a^2+bc=1=d^2+bc\implies a=\pm d\end{align}$$
But we also have that
$$b(a+d)=0=c(a+d)\implies \begin{cases}a=-d\\or\\b=c=0\end{cases}$$
In the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
When is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$? Let $x$, $y$, $z$ be three natural numbers such that $1< x< y< z$. For how many sets of values of $(x,y,z)$, is $(x-1)(y-1)(z-1)$ a factor of $xyz-1$?
I noticed that $(x-1)(y-1)(z-1)=(xyz-1)-z(x+y-1)-xy+x+y$.
But i don't know how to proceed from here. Any clues?
| There are no other solutions besides $(2,4,8)$ and $(3,5,15)$, already
discovered in aRaRa's answer.
Let $f(x,y,z)=\frac{xyz-1}{(x-1)(y-1)(z-1)}$ for integers $z>y>x>1$. We want
to know when $k=f(x,y,z)$ is an integer. Note first that $k$ cannot equal $1$,
because this would imply $z=\frac{x+y-xy}{x+y-1}$ whence $xy\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't g... | Given $x,y\ge 1$.
$\therefore x<\sqrt{x^2+x}$ and $y<\sqrt{y^2+y}$ $\implies$ $x+y<\sqrt{x^2+x}+\sqrt{y^2+y}$
$$\frac{x+y+1}{x+y}>\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} $$ and
$\frac{x+y+1}{x+y}=1+\frac{1}{x+y}\le \frac{3}{2}$, because $x+y\ge2$
$$\therefore \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} <2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
a way to integrate: $\int (\sqrt{x} +3)/(2+ x^ \frac{1}{3}) dx$ Im looking for a way to integrate:
$$
\int \frac{ \sqrt{x} +3}{2+ x^ \frac{1}{3}} dx
$$
that would make it efficient and not too difficult...
Any suggestions?
| Let $x=u^6$, $dx=6u^5du$.
\begin{align*}
&\int\frac{\sqrt x+3}{2+\sqrt[3]x}\,dx\\
=&\int\frac{u^3+3}{u^2+2}6u^5\,du\\
=&6\int\frac{u^8+3u^5}{u^2+2}\,du\\
=&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+\frac{12u+16}{u^2+2}\right)\,du\\
=&6\int\left(u^6-2u^4+3u^3+4u^2-6u-8+6\frac{2u}{u^2+2}+8\frac{1}{\left(\frac{\sqrt2}{2}u\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
System of recursive equation. Let's consider:
$$u_o = -1, v_0 = 3$$
$$\begin{cases} u_{n+1} = u_n + v_n \\ v_{n+1} = -u_n + 3v_n \end{cases}$$
I tried:
$$x^n = u_n , y^n = v_n$$
$$\begin{cases} x^{n+1} = x^n + y^n \\ y^{n+1} = -x^n + 3y^n \end{cases}$$
$$\begin{cases} x = 1 + \frac{y^n}{x^n} \\ y = -\frac{x^n}{y^n} +... | $$\begin{cases}
u_o = -1 \\
v_0 = 3 \\
u_{n+1} = u_n + v_n \\
v_{n+1} = -u_n + 3v_n
\end{cases}$$
So:
$$\begin{bmatrix}
u_{n+1} \\
v_{n+1} \\
\end{bmatrix} =
\begin{bmatrix}
1 & 1 \\
-1 & 3
\end{bmatrix}
\begin{bmatrix}
u_{n} \\
v_{n} \\
\end{bmatrix}$$
In order to make it nonrecursive, convert the sequence of matrix ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| Since $A+B+C=\pi$, we can write
$$
\sin^2 A=\sin^2(\pi-(B+C))=\sin^2(B+C)
$$
Expand with the sum formula:
$$
\sin^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B\sin^2C
$$
Transform $\sin^2$ into $\cos^2$:
$$
\cos^2C-\cos^2B\cos^2C+2\sin B\sin C\cos B\cos C+\cos^2B-\cos^2B\cos^2C
$$
Reduce and collect:
$$
\cos^2B+\cos^2C-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
calculate expected value of the product of two non independent random variables I've got two independent bernoulli distributed random variables $X$ and $Y$ with parameter $\frac{1}{2}$. Based on those I define two new random variables
$X' = X + Y , E(X') = 1$
$Y' = |X - Y|, E(Y') = \sum_{x=0}^1\sum_{y=0}^1|x-y|*P(X=x)... | Someone else can answer more authoritatively for the general case, but for a small experiment such as this one can we build up all possible values of $X' \cdot Y'$ from the four possible outcomes of $(X,Y)$?
$$
\begin{array}{l|l|l|l|l}
(X,Y) & X' & Y' & X' \cdot Y' & P(\ \ ) \\
\hline
(0,0) & 0 & 0 & 0 & \frac{1}{4} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How many ordered triples $(a, b, c)$ exist? How many ordered triples $(a, b, c)$ of positive integers exist with the property
that $abc = 500$?
Breaking it up, $500 = 2^2\cdot5^3$
$abc = 2^2 \cdot 5^3 = 2\cdot 2 \cdot 5 \cdot 5 \cdot 5$
But how can I use this?
| We solve a slightly more general problem, the number of triples $(a,b,c)$ of positive integers such that $abc=n=2^s5^t$. The solution generalizes nicely to $k$-tuples, and any positive $n$, given the prime power factorization of $n$.
The answer is the number of ways to distribute the $s$ $2$'s between $a$, $b$, and $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to go about solving this inequality question? $\cos(3x-\pi/3) \leq (1/2).$
Here is what I have done so far...
Let $3x-\pi/3 = X$. So I need to solve $\cos(X) \leq 1/2$. Which is all $X$ from $\pi/3$ to $5\pi/3$, so--
$\pi/3 \leq X \leq 5\pi/3 \quad\longrightarrow\quad \pi/3 \leq 3x-\pi/3 \leq 5\pi/3.$
So should I ... | It's more correct to say that
$$ \frac{\pi}{3} + 2n\pi \le 3x - \frac{\pi}{3} \le \frac{5\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} $$
Solving for $x$:
$$ \frac{2\pi}{3} + 2n\pi \le 3x \le 2\pi + 2n\pi $$
$$ \frac{2}{9} + \frac{2n\pi}{3} \le x \le \frac{2\pi}{3} + \frac{2n\pi}{3} $$
or
$$\frac{2 + 6n}{9}\,\pi \le x \le ... | {
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Is this logically valid? $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1} > ln(n)$$
and so, necessarily, $$1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n-1}+\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}.....+\frac{1}{n} > ln(n)$$
assuming that $n \in \mathbb {N}$.
| Imagine an inequality $$\frac{1}{n}>0.$$ We can add the same quantity on both sides $$\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)+\frac{1}{n}>0+\left(1+\frac{1}{2}+...+\frac{1}{n-1}\right)$$
Now we use your first inequality and the transitivity of $>$, which says that if $a>b$ and $b>c$, then $a>c$, where $a:=\left(1+... | {
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Prove that $2^{3^n} + 1$ is divisible by 9, for $n\ge1$
Prove that $2^{3^n} + 1$ can be divided by $9$ for $n\ge 1$.
Work of OP: The thing is I have no idea, everything I tried ended up on nothing.
Third party commentary: Standard ideas to attack such problems include induction and congruence arithmetic. (The answe... | Now, since we are trying to prove something for all n, we should look immediately to proof by induction. For $n = 1$, the result is trivial as $2^{3^1}+1 = 9$. Next, we assume that $2^{3^{(n-1)}} + 1$ is divisible by $9$, say $2^{3^{(n-1)}} + 1=9\cdot k$. So, $2^{3^{(n-1)}}=9\cdot k -1$. Now, we examine $2^{3^n}+ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$.
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong.
Can someone hel... | Following your way:
$$\begin{align}\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx &=8\int_0^{\arctan(1/2)}\tan^3\theta\sec\theta{\rm d}\theta\tag{$x=2\tan\theta$}\\&=8\int_1^{\sqrt{5}/2}(t^2-1){\rm d}t\tag{$t=\sec\theta$}\\&=8\left(\frac{t^3}3-t\right)\Bigg|_1^{\sqrt5/2}\\&=\frac83\left(\frac{5\sqrt5}8-1\right)-8\left(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to use U substitution for the integral $\int\frac{8}{49+x^2}\,dx$? So, the following is my problem.
$$\int\frac{8}{49+x^2}\,dx$$
I understand this. I should first take out the constant which is 8 so it'll be
$$8\int\frac{1}{49+x^2}\,dx$$
Then I should factor out the 49 no? So it'll be
$$\frac{8}{49}\int\frac{1}{1+... | Here are the steps
$$\int\frac{8}{x^2+49}dx= 8\int\frac{1}{x^2+7^2}dx
$$
$$=\frac{8}{7^2} \int\frac{1}{\frac{x^2}{7^2}+1}dx =\frac{8}{7^2} \int\frac{1}{\left(\frac{x}{7}\right)^2+1}dx $$
Let $u=\frac{x}{7}$, then $du=\frac{1}{7}dx$. So now
$$ \frac{8}{7} \int\frac{1}{u^2+1}du= \frac{8}{7} \arctan u+C $$
$$ = \frac{8}... | {
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Find all integers $x$ such that $x^2+3x+24$ is a perfect square. Find all integers $x$ such that $x^2+3x+24$ is a perfect square.
My attempt:
$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$
Now, do I find solution treating cases? But that doesn't seem very easy. Please help.
| Complete the square to get $(x+3/2)^2 + 87/4$. We want this to be a square itself, so
$$(x+3/2)^2 + 87/4 = k^2.$$
This is the same as
$$(2x+3)^2 + 87 = 4k^2.$$
Now consider the difference of squares $(2k - 2x - 3)(2k + 2x + 3) = 87 = 3\cdot 29$. Now there are only a few cases to check.
*
*$2k + 2x + 3 = 3$ and $2k ... | {
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$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to vari... | Just knowing that $a^2+b^2+c^2=1$ is not enough to determine the value of $ab+bc+ca$. For example, if $a=b=0$ and $c=1$, then $ab+bc+ca = 0$. On the other hand, if $a=b=c=\frac{1}{\sqrt3}$, then $ab+bc+ca = 1$. In fact, using $$(a+b+c)^2=a^2+b^2+c^2 + 2(ab+bc+ca),$$ you get that $$ab+bc+ca=\frac{(a+b+c)^2 - 1}{2}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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How to derive this infinite product formula? Show:
$$\prod_{n=0}^{\infty}\left(1 + x^{2^n}\right) = \frac{1}{1-x}$$
I tried numerous things, multiplying by $x$, dividing, but none of that worked. Also, I realized that:
$$\prod_{n=0}^{\infty} \left(1 + x^{2^n}\right) = \sum_{n=0}^{\infty} x^n$$
But I cannot prove the r... | The partial products are
$$1+x,$$
$$(1+x)+x^2(1+x)=1+x+x^2+x^3,$$
$$(1+x+x^2+x^3)+x^4(1+x+x^2+x^3)=1+x+x^2+x^3+x^4+x^4+x^5+x^6+x^7$$
$$\cdots$$
Every time the number of terms doubles.
Doesn't that ring a bell ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x}... | We have
$$x\log(x) = x[-\log(1/x)] = x\left[-\log\left(1 - \frac{x-1}{x}\right)\right] = x\sum_{m = 0}^\infty \frac{1}{m+1}\left(\frac{x-1}{x}\right)^{m+1},$$
and $$x = \dfrac{1}{\frac{1}{x}} = \dfrac{1}{1 - \frac{x-1}{x}} = \sum_{m = 0}^\infty \left(\frac{x - 1}{x}\right)^m.$$ Thus
$$x\log(x) = \sum_{m = 0}^\infty \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130883",
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"source": "stackexchange",
"question_score": "5",
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Find coefficient of x in a generating function The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
I factored out some $x$'s, to get $x^3(1+x^2+x^3)x^4(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$and then combined the factored terms to get $... | From the sum $\frac{1}{1-x^5}$ you only need to consider the first two terms - other will give you a higher power. $10 = x_1 +x_2+ x_3$ (with terms all three sums). Can you idenify $x_k$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Matrix $\mathbf C$ such that $\mathbf A = \mathbf B \mathbf C=\mathbf C\mathbf B$ Let
$$\mathbf A =\begin{pmatrix}1&1&1\\1&2&2\\1&2&3\end{pmatrix} \text{and }\mathbf B=\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}$$
Then which of following is true
*
*There exists a matrix $\mathbf C$ such that $\mathbf A=\mathbf ... | I consider $$C=\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i \end{array}\right),$$ then
$$BC=\left(\begin{array}{ccc}a&b&c\\a+d&b+e&c+f\\a+d+g&b+e+h&c+f+i \end{array}\right),$$
and
$$CB=\left(\begin{array}{ccc}a+b+c&b+c&c\\d+e+f&e+f&f\\a+d+g&h+i&i \end{array}\right).$$
If, we want $A=BC$, we get
$$C=\left(\begin{array}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$
My attempt:
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{... | I think you can do this using Lagrange multipliers. Define the function:
$$f(x,y,z)= \frac{x^3}{(1+y)(1+z)} +\frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} - \frac{3}{4}.$$
We want to show that $f(x,y,z)\geq 0$ for all $x,y,z\in \mathbb R^3$ such that $xyz=1$. Define the Lagrangian:
$$L(x,y,z,\lambda) = f(x,y,z)- \la... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having trouble solving part two of an equation: Part one consisted of proving that
$$\frac{x-1}{x-3} = 1+ \frac{2}{x-3}$$
I completed this and here is my working:
$$let \frac{x-1}{x-3}= LHS$$
$$RHS=\frac{x-3}{x-3} + \frac{2}{x-3}$$
$$\frac{2+(x-3)}{(x-3)}$$
$$\frac {x-3+2}{x-3}$$
$$\frac {x-1}{x-3} = LHS$$
Part two th... | Start by realizing you can get a common denominator (and simplify once you have):
$$
-\frac{4}{(x-5)(x-3)}=-\frac{4}{(x-9)(x-7)}.
$$
Now take the reciprocal of both sides:
$$
-\frac{1}{4}(x-5)(x-3)=-\frac{1}{4}(x-9)(x-7).
$$
Now expand out the terms on the left- and right-hand sides:
$$
-\frac{x^2}{4}+2x-\frac{15}{4}=-... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve $\sum_{n=5}^\infty \frac{12}{16 \, n^{2} + 40 \, n + 21}$ How do I solve
$\displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}$ ?
A tip was to turn this into a telescoping series but I don't see how I could do this. Are there any guidelines in how I could transform this seri... | $\begin{align} \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=\displaystyle\frac{12}{(4n+3)(4n+7)}\\&=3\frac{4}{(4n+3)(4n+7)}\\&=\displaystyle3{\huge[}\frac{1}{(4n+3)}-\frac{1}{(4n+7)}{\huge]}\end{align}$
$\begin{align}\therefore \displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ Find:
$$ \sum_{n=2}^{+\infty}\frac{1}{n^{3}-n}$$
I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks
| We know that the series at hand is convergent absolutely, so we can safely play with the sum.
$$\sum_{i=2}^\infty\frac{1}{n^3 - n} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=2}^\infty\frac{1}{n+1} + 0.5\sum_{i=2}^\infty\frac{1}{n-1} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=3}^\infty\frac{1}{n} + 0.5\sum_{i=1}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $n^2+4n+3$ is not prime for $n \in \mathbb{Z}^{+}$. I am trying to write a proof for this theorem:
For every positive integer $n$, $n^2+4n+3$ is not a prime.
Proof: Let $n \in \mathbb{Z}^{+}$. Note that $$n^2+4n+3=(n+1)(n+3)>1\text{,}$$
and $n+1 >1$ and $n+3 >1$.
Let $a = n+1$ and $b = n+3$. Then we have $$\df... | One easy way of going about this is simply to consider the parity of $n$:
$n$ is even: We have that
$$
(2\ell)^2+4(2\ell)+3=4\ell^2+8\ell+3=\underbrace{(2\ell+1)(2\ell+3)}_{\text{composite}}.
$$
Thus, $n^2+4n+3$ is not a prime when $n$ is even.
$n$ is odd: We have that
$$
(2\ell+1)^2+4(2\ell+1)+3=\underbrace{2(2\ell^2+... | {
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"url": "https://math.stackexchange.com/questions/1143190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplify the sum Consider the sum of two polynomials
$$
\sum_{k=0}^{n-1} {{{n-1} \choose {k}}^2 z^{2k}}+\sum_{k=0}^{n-2} {{n-2} \choose {k}} {{n} \choose {k+1}} z^{2k+1}=\sum_{i=0}^{2n-1}a_i z^i.
$$
I want to find the exact expression for $a_i$. Of course I may divide it into even and odd parts but I hope there is a ... |
A sum $A(z)=\sum_{k=0}^{2n}a_kz^k$ can be split into even and odd parts via
\begin{align*}
\frac{1}{2}\left(A(z)+A(-z)\right)
&=\frac{1}{2}\left(\sum_{k=0}^{2n}a_kz^k+\sum_{k=0}^{2n}a_k(-z)^k\right)\\
&=\sum_{k=0}^{2n}\frac{1+(-1)^k}{2}a_kz^k\\
&=\sum_{k=0}^na_{2k}z^{2k}
\end{align*}
and
\begin{align*}
\frac{1}{... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculating the limit: $\lim \limits_{x \to 0}$ $\frac{\ln(\frac{\sin x}{x})}{x^2}. $ How do I calculate $$\lim \limits_{x \to 0} \dfrac{\ln\left(\dfrac{\sin x}{x}\right)}{x^2}\text{?}$$
I thought about using L'Hôpital's rule, applying on "$\frac00$," but then I thought about $\frac{\sin x}{x}$ which is inside the $\... | Using taylor series you have
$$\color{#05f}{\frac{\sin x}{x}} = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n}}{(2n+1)!} = 1 - \frac{x^2}{6} + \frac{x^4}{120} + O(x^6)$$
and $$\ln \Bigg(\color{#05f}{\frac{\sin x}{x}}\Bigg) = - \frac{x^2}{6} - O(x^4)$$
$$\lim _{x\to 0 } \frac{- \frac{x^2}{6} - O(x^4)}{x^2}= \lim_{x\to 0} -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Simplify a series of $\cos x$ or $\sin x$ Suppose we have this series
$$\sum\limits_{k=0}^n a^k \cos(kx) = 1 + a\cos(x) + a^2 \cos(2x) + a^3 \cos(3x) + \cdots + a^n\cos(nx)$$
What should we do to simplify this?
| let $a$ be a real number and $z = \cos t + i \sin t$.
this looks like the real part of $$1 + az + (az)^2 + \dots + (az)^n = \frac{1- (az)^{n+1}}{1-az} $$ thanking the real part, we get
$$1 +a\cos t + a^2 \cos 2t + \dots+a^n \cos nt = \text{real part} \frac{1- (az)^{n+1}}{1-az}$$
finding the real part of $$\frac{1- (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147717",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If $f(x) = 3x^2-x + 2$, find $f(a)$ and $[f(a)]^2$ If $f(x) = 3x^2-x^2$, find $f(a)$ and $[f(a)]^2$
Also, $2f(a) = 3x2(a)^2-2(a)+2 = 6a^2-2a+2$
The book says the answer is $6a^2-2a+4$. Why is that? Is the book wrong?
| what you are computing in your work sheet is $f(f(a).$ that is usually is different from $[f(a)]^2.$ you were asked to compute $[f(a)]^2 = (3a^2 - a + 2)^2 = 9a^4 + \dots +4.$
| {
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"timestamp": "2023-03-29T00:00:00",
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inequalites of an acute triangle angles $ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $ If $a,b,c$ are an acute angle of triangle the prove that
$ 180^{180}*a^b*b^c*c^a \le (a^2+b^2+c^2)^{180} $
No idea
| Taking $180^{\text{th}}$ roots and using $a+b+c=180^\circ$, the inequality is equivalent to
$$(a^b b^c c^a)^{\frac1{a+b+c}} \le \frac{a^2+b^2+c^2}{a+b+c}$$
Use weighted AM-GM in the form
$$(a^b b^c c^a)^{\frac1{a+b+c}} \le \frac{b\cdot a + c\cdot b+a\cdot c}{b+c+a}$$
and $ab+bc+ca \le a^2+b^2+c^2$ to conclude.
| {
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Maximisation: Half circle area inscribed within isoceles triangle? Given an isosceles triangle with the line of symmetry along $x=0$, and the odd side along $y=0$, How can I optimize the maximum parabolic area inscribed within the triangle? The "half -circular" area has points of tangency to the triangle, one of each s... | Let the triangle have side with equation $y_1=c-mx$. Start here.
Let the parabola have equation $y_2=b-ax^2$.
Gradients are $\dfrac{dy_1}{dx}=-m$ and $\dfrac{dy_2}{dx}=-2ax$
Gradients equal when $m=2ax$, this is ${x}=\dfrac{m}{2a}$.
Must also have same $y$ value, so
$$c-m\frac{m}{2a}=b-a\frac{m}{2a}\frac{m}{2a}$$
$$c-\... | {
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"url": "https://math.stackexchange.com/questions/1151283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the probability that each of the 8 members serves on at least one of the three committees. A certain group has 8 members. In January, 3 members are selected at random to serve on a committee. In February, 4 members are selected at random and independently of the first selection to serve on another committee.... | WLOG, assume that the members $1$ to $5$ belong to the third committee.
The probability that $4$ of those members belong to the second committee, is
$\frac{\binom{5}{4}}{\binom{8}{4}}=\frac{1}{14}$. In this case, all the remaining
members must belong to the first committee, so $\frac{1}{14}\times\frac{1}{\binom{8}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151363",
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"source": "stackexchange",
"question_score": "2",
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How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the ri... | Check that the equation holds for $n=2$. Assuming that it holds for some $n$:
$$\begin{align}\prod_{i=2}^n\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\\
\end{align}$$
And so
$$\begin{align}
\prod_{i=2}^{n+1}\left( 1-\frac1{i^2}\right)&=\frac{n+1}{2n}\left( 1-\frac1{(n+1)^2}\right)\\
&=\frac{n+1}{2n}-\frac{n+1}{2n(n+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^... | $$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x.$$
Setting $x=-4$,
$$0A+0B+18C=-20.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle. Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.
| Doing this in $\mathbb{R}^3$, let the three points of the equilateral triangular base of the tetrahedron be $(0,0,0)$, $(0,2\ell,0)$, and $(\sqrt{3}\ell,a,0)$. Suppose the lengths of the edges of the tetrahedron between each of these point and our apex are $a$, $b$, and $c$ respectively. The apex (if it exists) will be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method t... | Hint: $x^2 \equiv -1$ mod $5$ $\iff$ $x^2 \equiv 4$ mod $5$ $\iff$ $(x-2)(x+2) \equiv 0$ mod $5$, whence $x \equiv 2$ or $x \equiv -2$ mod $5$.
$x^2 \equiv -1$ mod $13$ $\iff$ $x^2 \equiv 25$ mod $13$ $\iff$ $(x-5)(x+5) \equiv 0$ mod $13$, whence $x \equiv 5$ or $x \equiv -5$ mod $13$.
Now cross-solve these $4$ possib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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if $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$, then $x\in\mathbb{Z}$. Assume that $x^3-x\in\mathbb{Z}$ and $x^4-x\in\mathbb{Z}$ for some $x\in\mathbb{R}$.
Prove that $x\in\mathbb{Z}$.
my attempt:
Let $a=x^3-x$ and consider polynomial $X^3-X-a$, then $x$ is a root of it and if $x\in\mathbb{Q... | Let $x^3-x=a$ and $x^4-x=b$.
Let's do a few steps of the Euclidean algorithm for polynomial GCD:
$x^4-x-b=x(x^3-x-a)+(x^2+(a-1)x-b)\\ \implies x^2+(a-1)x-b=0$
$x^3-x-a=x(x^2+(a-1)x-b)+((1-a)x^2+(b-1)x+a)\\ \implies (1-a)x^2+(b-1)x+a=0$
$0=(1-a)(x^2+(a-1)x-b)-((1-a)x^2+(b-1)x+a)$
$\implies x ((1-a) (a-1)-b+1)-(1-a) b-a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find : $ dy/dx, y=\sqrt{4x^2 - 7x - 2}$ The problem says Find $dy/dx, y=\sqrt{4x^2 - 7x - 2}$
So far I changed it to $(4x^2 - 7x - 2)^{1/2}$
I don't know where to go from there.
| If $y = \sqrt{4x^2 - 7x - 2}$
then
$$\frac{dy}{dx} = \frac{d}{dx}(4x^2 - 7x - 2)^{\frac{1}{2}}
\\
$$
$$
=\frac{1}{2}\frac{d}{dx}(4x^2 - 7x - 2) \times (4x^2 - 7x - 2)^{\frac{-1}{2}}
$$
$$
= \frac{8x-7}{2\sqrt{4x^2 - 7x - 2}}
$$
using the chain rule.
That is, if we have a function $y = u^{\frac{1}{2}}$ then $$\frac{dy}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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solving cubic diophantine equation Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.
My Algebraic Approach:
$x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.
The only... | To bound all solutions
in integers
(not just to positive integers)
to
$x^3+y^3 = n$:
First,
using the factorization
$x^3+y^3
=(x+y)(x^2-xy+y^2)
$,
we get
possible values
for $x+y$
since $(x+y) | n$.
Then,
since
$x^2-xy+y^2
=(x+y)^2-3xy
$,
we get possible values
for $xy$
($xy
=\dfrac{(x+y)^2-n/(x+y)}{3}
$)
and
this give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum $\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$ Find the value of
$$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$$
I have tried to substitute the value of $n$ from $1$ but it is not telescoping series.
| Hint. Recall the following series representation for the digamma function
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant.
Then by partial fraction decomposition we have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim\limits_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$ I don't know how to evaluate
$$\lim_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$$
*
*It is $\frac{\infty}{\infty}$
*I have tried to multiply by $5 - \sqrt{x^2+5}$, but I reach a wrong result.
*I have tried to take the derivative of $5 + \sqrt{x... | Derivative of the numerator: $(5+ \sqrt{x^2+5})'= (\sqrt{x^2+5})'= \frac{(x^2+5)'}{2}\sqrt{x^2+5}^{-1} = \frac{2x}{2 \sqrt{x^2+5}} = \frac{1}{\sqrt{1+\frac{5}{x^2}}}$.
Derivative of the denominator: $(x-6)'=1$.
L'Hospital's rule is possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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critical points of the function Find the critical points of the function
$$f(x,y)=(4x-x^2)\cos y$$
first let's determinate Partial derivatives:
$$\dfrac{\partial f}{\partial x}(x,y)=(4-2x)\cos y$$
$$\dfrac{\partial f}{\partial y}(x,y)=-\sin y(4x-x^2) $$
To find the critical points, we solve:
\begin{cases}\dfrac{\partia... | Hint: The second equation is $x(x-4)\sin y=0$ which gives $x=0$ or $x=4$ or $y=k\pi$, $k \in \Bbb{Z}$.
Consider all three cases separately by plugging them in to the first equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate a rational function of $x,y,z$ given two polynomial equations in $x,y,z$ Let $x, y, z$ be real numbers. Given that $$2x(y^2−1)+2y(x^2−1)=(1+x^2)(1+y^2)$$ and $$4z(1−y^2)+4y(1−z^2)=(1+z^2)(1+y^2)$$ Find the value of the following expression:
$$\Bigg(\frac{2x}{1+x^2}−\frac{2z}{1+z^2}\Bigg)^2+\Bigg(\frac{1−z^2}{1... | I don't know if a better solution is possible, but here is mine.Consider the following points:
$A(\frac{2x}{1+x^2},\frac{1−x^2}{1+x^2}),B(\frac{1−y^2}{1+y^2},\frac{2y}{1+y^2}),C(\frac{2z}{1+z^2},\frac{1−z^2}{1+z^2})$.
Notice that these points on a unit circle, as for any real number $a$,
$$(\frac{2a}{1+a^2})^2+(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
Find all ordered tuples of positive integers $(a_1,a_2,a_3,\ldots,a_n)$ such that
$\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
The only... | Partial solution: If $n+1$ is a square, then set $a_1=a_2=\cdots=a_n=\sqrt{n+1}$.
Then the LHS is $$(1+2+\cdots+n)\frac{1}{\sqrt{n+1}}=\frac{n(n+1)}{2\sqrt{n+1}}=\frac{1}{2}n\sqrt{n+1}$$
which agrees with the RHS.
More of a partial solution, following @Stefan4024's induction idea. Start with a solution of the prece... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
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Floor function right hand limit Show
$$\lim_{x\to 0^+}\frac{x}a\cdot\left\lfloor\frac{b}x\right\rfloor=\frac{b}a\;.$$
I think I should use boundedness of $\left\lfloor\dfrac{b}x\right\rfloor$, $$\frac{b}x-1\le\left\lfloor\frac{b}x\right\rfloor\le\frac{b}x\;.$$
But I am not sure how to proceed with this.
| As you mention, $|\lfloor b/x \rfloor - b/x| \leq 1$, and so
$$
\frac{x}{a} \left|\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{x}\right| \leq \frac{x}{a}.
$$
We can simplify this to
$$
\left|\frac{x}{a}\left\lfloor \frac{b}{x} \right\rfloor - \frac{b}{a}\right| \leq \frac{x}{a}.
$$
As $x\to0^+$, the right-hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find lim sup lim inf How do I find $\lim \sup\text{ or } \lim \inf$ of $ \sin (\frac{n\pi}{5})$ ?
I know the $\sin$ function normally oscillates between $-1$ and $1$ but that obviously is not the answer for $\lim \inf$ and $\lim \sup$.
| Observe that $$\begin{align*}&\left\{ \sin \left(\frac{n\pi}{5}\right)|n\in \mathbb{N}\right\}\\&=\left\{\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),\sin \left(\frac{5\pi}{5}\right),\sin \left(\frac{6\pi}{5}\right),\sin \left(\frac{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
| $$n^5+29n-(n-2)(n-1)n(n+1)(n+2)=5n^3+25n=5(n-1)n(n+1)+30n$$
$$\implies n^5+29n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{The product of }5\text{ consecutive integers}}+5\underbrace{((n-1)n(n+1))}_{\text{The product of }3\text{ consecutive integers}}+30n$$
See The product of n consecutive integers is divisible by n fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
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If $ f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then real solution of $f(f(x))=0$
If $\displaystyle f(x) = \left\{\frac{3x}{2}\right\}\;,$ Where $\{x\}=x-\lfloor x \rfloor$. Then no. of solution of the equation
$f(f(x))=0$ and $f(f(f(x)))=0$ and $f(f(f(f(x))))=0$.
$\bf{My\; Try::}$ We ... | Note that $f(f((\frac23)^2))=0$, so $0$ is not the only solution.
$\bullet$ If $0\leq x<\displaystyle \frac{2}{3} .$ Then we use $\displaystyle f(x) = \frac{3x}{2}.$ So $\displaystyle f(f(x)) = 0\Rightarrow \frac{9x}{4}=0\Rightarrow x=0$..
The mistake you are making is that even when $0\le x\le \frac23$, $f(x)$ may n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$
$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$, where $\left(\frac{.}{.}\right)$ is the Jacobi symbol.
For example I know that $\left(\frac{2}{15}\right)=\left(\frac{2}{... | For the second part of your question which remains unanswered, one can find all the quadratic residues $\pmod n$ by considering all numbers up to $n$ modulo $n$, squaring all of them and considering the residues $\pmod n$. Notice that any other numbers, for example $n+2$, would have the same quadratic residue as the nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Does this qualify as a Laurent series? Evaluate the Laurent series around the singularity at $ z_0 = 3$.
$$ \frac{1}{z^2(z-3)} $$
I can apply the geometric series as follows:
$$\frac{1}{z^2}\cdot\frac{1}{z-3}=-\frac{1}{3z^2}\cdot\frac{1}{1+\frac{z}{3}}=\frac{1}{3z^2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z}{3}\right)^n$... | For a Laurent series about the point $z_0=3$, you want the series to be in powers of $z-3$.
$$
\begin{align}
\frac1{z^2(z-3)}
&=\frac1{((z-3)+3)^2(z-3)}\\
&=\frac1{9\left(1+\frac{z-3}3\right)^2}\cdot\frac1{z-3}\\
&=\frac19\frac1{z-3}\left(1-2\frac{z-3}3+3\frac{(z-3)^2}9-4\frac{(z-3)^3}{27}+\dots\right)\\
&=\frac1{9(z-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this inequality with $xyz\le 1$
If $x,y,z\ge 0$ and $\color{red}{xyz\le 1}$, show that
$$\color{blue}{\dfrac{x^2-x+1}{x^2+y^2+1}+\dfrac{y^2-y+1}{y^2+z^2+1}
+\dfrac{z^2-z+1}{z^2+x^2+1}\ge 1}.$$
| Multiplying and dividing each of the expressions numerator by $(1 +
x)$, $(1 + y)$ and $(1 + z)$ respectively:
$$
\frac{z^3+1}{(z+1) \left(x^2+z^2+1\right)}+\frac{x^3+1}{(x+1) \left(x^2+y^2+1\right)}+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}
$$
Further simplifying:
$$
1+ \frac{x^2-x+1}{x^2+y^2+1}-\frac{x^2+z}{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$ How to prove that $$\displaystyle\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$$
Thanks in advances.
| Rewrite a partial product as
$$\prod_{n=0}^N \frac{(4 n+2)^3 (4 n+4)}{(4 n+1)(4 n+2)(4 n+3) (4 n+4)} = \frac{2^{3 N+3} (2 N+1)!!^3 4^{N+1} (N+1)!}{(4 N+4)!} $$
Use the fact that
$$(2 N+1)!! = \frac{(2 N+1)!}{2^N N!} $$
and
$$M! \approx M^M e^{-M} \sqrt{2 \pi M} \quad (M \to \infty)$$
The rest is careful bookkeeping, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Integral solutions to inverse trigonometric equation. The original question is this : I have to find number of ordered pairs of integral solutions to this :
$$\tan^{-1}{x} + \cos^{-1}{\frac{y}{\sqrt{1+y^2}}} = \sin^{-1}{\frac{3}{\sqrt{10}}} $$
I rearranged this as $\tan^{-1}(\frac{1}{y}) = tan^{-1}{3} - \tan^{-1}x $
An... | $\bf{My\; Solution::}$ Given $$\displaystyle \arctan x + \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arcsin \left(\frac{3}{\sqrt{10}}\right)$$
Now We can Write $$\displaystyle \arccos\left(\frac{y}{\sqrt{1+y^2}}\right) = \arctan\left(\frac{1}{y}\right)$$ and $$\displaystyle \arcsin\left(\frac{3}{\sqrt{10}}\right)=\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving $x^2 (x-y)^2 = x^2 - y^2$ using implicit differentiation I have a test tomorrow and I am doing homework to review and study. The problem is to differentiate
$$
x^2 (x-y)^2 = x^2 - y^2.
$$
I tried multiple times; however, every time I try I get the incorrect answer. It would be greatly appreciated if somebody co... | $\frac{d}{dx}[x^2(x-y)^2 = x^2 - y^2]$
$\Rightarrow 2x(x-y)^2 + x^2\cdot 2(x-y) \cdot (1-\frac{dy}{dx}) = 2x - 2y\cdot\frac{dy}{dx}$
$\Rightarrow 2x^3-4x^2y + 2xy^2 +2x -2x\frac{dy}{dx}-2y + 2y\frac{dy}{dx}= 2x -2y\frac{dy}{dx}$
$\Rightarrow 2x^3 - 4x^2y + 2xy^2 - 2y = (2x-4y)\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{x^3 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Problem in trigonometry integral $$\int \frac{1}{(9-x^2)^{\frac{3}{2}}}dx$$
Let $x=3\sin (u)$
$dx=3\cos (u)$
So, $(9-x^2)^{\frac{3}{2}}=(9-9\sin ^2(u))^{\frac{3}{2}}$
$=27\cos ^3(u)$
$u=\sin ^-1(\frac{x}{3})$
My problem is how to substitute u into the integral to become $3\int \frac{\sec ^2(u)}{27}$du?
| Using your substitutions:
$$
\int \frac{dx}{(9 - x^2)^{\frac{3}{2}}} \;\; =\;\; \int \frac{3 \cos u du}{27(1- \sin^2u)^{\frac{3}{2}}} \;\; =\;\; \frac{1}{9} \int\frac{\cos u}{\cos^3u} du \;\; =\;\; \frac{1}{9} \int \frac{du}{\cos^2u}.
$$
Now use the fact that $\sec u = \frac{1}{\cos u}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Series of inverse function $A(s) = \sum_{k>0}a_ks^k$ and $A(s)+A(s)^3=s$.
I want calculate $a_5$. What ways to do it most efficiently?
| We can elaborate on the Lagrange inversion concept.
Suppose we have $$A(s) = \sum_{n\ge 0} a_n s^n$$
and $A(s)+A(s)^3=s$ and we seek $a_n.$
Using the Cauchy Residue Theorem to prepare for Lagrange inversion we have that
$$a_n =
\frac{1}{2\pi i}
\int_{|s|=\epsilon}
\frac{1}{s^{n+1}} A(s) \; ds.$$
Now put $A(s)=w$ so t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If Matrix $A^3 = 0$ then what will be $I+A+A^2$ Matrix $A$ is a square matrix such that $$A^3=0$$ then what would be value of $$I+A+A^2=?$$ such that $I$ is unit matrix of same order as $A$
I firstly supposed sum to be $X$ that is
$X=I+A+A^2$ then
$$AX=A+A^2+A^3$$
$$AX=A+A^2$$
and multiplying both sides with $A^{-1}$... | If $A^3=0$, then $A^4=0$, $A^5=0$, and so on, so $I+A+A^2=I+A+A^2+A^3+A^4+\cdots$. The infinite sum $I+A+A^2+A^3+A^4+\cdots$ looks a lot like an infinite geometric series. There’s a formula for the sum of an infinite geometric series (of real numbers, if the ratio has absolute value less than $1$), $\frac{a}{1-r}=a+ar+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Solving $3\sin x - \cos x = 2$ for $x \in [0, 2\pi)$ Problem:
Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$
My attempt:
I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.
I've... | Another approach is to write: $$A \sin(x) + B \cos(x)=C$$ and then divide by $$\sqrt{A^2 + B^2}$$ to get:
$$\frac{A}{\sqrt{A^2+B^2}}\sin(x) + \frac{B}{\sqrt{A^2+B^2}}\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$
Since $$\left(\frac{A}{\sqrt{A^2+B^2}}\right)^2 + \left(\frac{B}{\sqrt{A^2+B^2}}\right)^2 = 1$$ there is a $\psi$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Find $y'$ for $\ln(x+y)=\arctan(xy)$
Find $y'$ for $\ln(x+y)=\arctan(xy)$
Here is my attempt at a solution. Is this correct? Any hints or advice would be appreciated.
| Even if the result is correct, I guess that a systematic use of total differentiation makes things slightly easier.
Consider $$F=\log(x+y)-\tan^{-1}(xy)=0$$ So $F'_x~dx+F'_y~dy=0$. I this case, $$F'_x=\frac{1}{x+y}-\frac{y}{x^2 y^2+1}$$ $$F'_y=\frac{1}{x+y}-\frac{x}{x^2 y^2+1}$$ $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Problem in Differential Equation (How to proceed?) $$y\frac{dy}{dx}=\sqrt{1-y^2}$$, y=0 when x=0
$$\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}=1$$
$$\int\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}dx=\int 1dx$$
$$-\sqrt{1-y^2}=x+c_1$$
$$1-y^2=(x+c_1)^2$$
$$y^2=1-(x+c_1)^2$$
$$y=\pm \sqrt{1-(x+c_1)^2}$$
When x=0, y=0
$$0=\pm \sqrt{1-c^2... | Square both sides:
$$1-y^2=(x+c_1)^2$$
rearrange for $y^2$:
$$y^2=1-(x+c_1)^2$$
take $\pm$ square root:
$$y=\pm\sqrt{1-(x+c_1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $p,q$ s.t. $2q^2-p^2=\Box$ and $2p^2-q^2=\Box$ Problem. Find all integers $p,q$ such that $2q^2-p^2$ and $2p^2-q^2$ are perfect squares.
I think this is only true when $p=\pm q$ but I have not been able to prove it.
One approach I tried is letting (wlog) $q=p+t$ with $t>0$ to get $p^2+4pt+2t^2$ and $p^2-2pt-t^2$ a... | We may assume WLOG $p,q\in \mathbb{N}^+$ and $\gcd(p,q)=1$.
Lemma 1. If $\gcd(a,b)=1$ and $a^2+b^2=2c^2$, then:
$$ \{a,b\} = \{u^2-2uv-v^2,u^2+2uv-v^2\},\quad c=u^2+v^2,\quad \gcd(u,v)=1. $$
Proof. We have that $P=\left(\frac{a}{c},\frac{b}{c}\right)$ is a rational point on the circle $\Gamma:x^2+y^2=2$, hence the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Mathematical Induction Proof Question dealing with integers How would you use mathematical induction to prove that
$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n \cdot (n + 1) \cdot (n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$
I tried proving the base case of $n = 1$ but the left half is much larger than the right ... | The base case for $n=1$ is actually quite easy, since:
$\frac{1\cdot 2\cdot 3 \cdot 4}{4} = 1\cdot 2 \cdot 3$
Assuming that it holds for $n$, we check $n+1$:
$1\cdot 2\cdot 3 + \cdots + n(n+1)(n+2) + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)}{4} = \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4 \sqrt{2014}$ How to prove that $\sqrt{2011} + \sqrt{2013} + \sqrt{2015} + \sqrt{2017} < 4\sqrt{2014}$ without using calculator?
| The function $f(x)=\sqrt{x}$ is strictly concave for $x>0$ so by Jensen's inequality, we have
$$
\frac{1}{4}f(2011)+\frac{1}{4}f(2013)+\frac{1}{4}f(2015)+\frac{1}{4}f(2017)<f\left(\frac{1}{4}(2011+2013+2015+2017)\right)
$$
which is $\sqrt{2014}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and ... | We have $$\frac{x^2y^2}{x^2+y^2}=\frac{1}{\dfrac{1}{x^2}+\dfrac{1}{y^2}}.$$ Note that each term in the denominator on the RHS is positive and independently goes to $\infty$ as $x$ and $y$ approach $0$. So the required limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Steady state solution for wave equation with gravity For the following wave equation and initial condition where G is a constant due to gravity, how would I go about finding the steady state solution:
$\frac{∂^2u}{∂t^2}= c \frac{∂^2u}{∂x^2}+ G, \quad 0\le x \le L,\: t\gt 0 $
$u(0, t) = 0 ,\quad t > 0,$
$u(L, t) = H ,\... | For the equation $u_{tt} = c^{2} u_{xx} + a$ with the conditions
\begin{align}
u(0, t) &= 0 \hspace{10mm} u(L, t) = H \\
u(x,0) &= 0 \hspace{10mm} u_{t}(x,0) = 0
\end{align}
consider a solution of the form $u(x,t) = F(x) G(t) + f(x)$, where $f(x)$ is the steady-state solution, for which
\begin{align}
u_{xx} &= F'' G +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?
| A simple way to see if a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested is to check if $a^2-b$ is a perfect square. In this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
(you can easely verify this identity).
In this case $a^2-b=35$ is not a perfe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluate:$\int_{0}^{\pi}\frac{x\sin2x\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$ Here i have a problem
$$\int_{0}^{\pi}\frac{x\sin (2x)\sin(\frac{\pi}{2}\cos x)}{2x-\pi}dx$$
I have started applying the property of definite integral but stucked after going few steps. Please help.
| Let $A$ denote the value of the integral. First note that since $\sin(2x)/(2x - \pi) = -\sin(2x - \pi)/(2x - \pi) \to -1$ as $x \to \pi/2$, the integral is proper. By a change of variable $u = 2x - \pi$, we have
$$A = \int_{-\pi}^\pi \frac{u + \pi}{2u}\cdot \left[-\sin u \sin\left(\frac{\pi}{2}\cos\frac{\pi + u}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove, that every $p$ prime has a multiple, which is smaller than $\frac{p^4}{4}$, it can be written down as the sum of five integer's fourth power. I have a really interesting (and hard) number theory task:
Prove, that every $p$ prime has a multiple(not $0$), which is smaller than $\frac{p^4}{4}$, and it can be writte... | As a first step, it is easy to see that the multiple $6p^2$ can be written as the sum of $12$ fourth powers. The reason is, that every integer of the form $6x^2$ can be written this way, because $x=a^2+b^2+c^2+d^2$ is the sum of four squares, and
\begin{align*}
6x^2 & =6(a^2+b^2+c^2+d^2)^2 \\
& =(a+b)^4+(a-b)^4+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the area of trapezium $ABCD$ is a trapezium in which $AB||CD$. If $P$ is the point of intersection of diagonals $AC$ and $BD$ such that area of triangle $DPC=50cm^2$ and area of triangle $APB=32cm^2$.Then find area of trapezium $ABCD$.
I found that triangle APB and DPC are similar.The ratio of their sides is 4:5.T... | Consider the diagram below:
We are given that $\overline{AB} \parallel \overline{CD}$. If two parallel lines are cut by a transversal, then alternate interior angles are congruent. Thus, $\angle ABP \cong \angle CDP$ and $\angle CAB \cong \angle DCP$. Therefore, $\triangle ABP \sim \triangle CDP$.
We are also giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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polar coordinates and complex numbers Prove that $\dfrac{-1+i\sqrt{3}}{2}$ is a cube root of $1$.
I believe I must use polar coordinates to solve this. Perhaps $z=r\cos(\theta)+i\sin (\theta)$. Any help would be great!
| Hint:
$x$ is a cubic root of $y$ if and only if $x^3 = y$. So see what $x^3$ is. No need to use what you call polar coordinates.
EDIT: $$ \left({\dfrac{-1+i\sqrt{3}}{2}}\right)^3 = \dfrac{(-1)^3 + 3(-1)^2( i\sqrt{3} ) + 3(-1)( i\sqrt{3})^2 + (i\sqrt{3})^3}{8} = \dfrac{-1 + 3 \sqrt 3 \cdot i + 9 - 3 \sqrt 3 \cdot i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$... | Alternatively, isolating $ \sqrt{4x-1}$ and then multiplying both sides by $\sqrt{x+1}+\sqrt{x-1}$ makes it easier to conclude the LHS is smaller than the RHS: $$\require\cancel \cancel{x}+1-\cancel{x}+1=2<\sqrt{4x-1}\left(\sqrt{x+1}+\sqrt{x-1}\right).\tag{$\star$}$$
Once we check $(\star)$ holds for $x=1$ we are done,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integrate $dx/(a^2 \cos^2x+b^2 \sin^2x)^2$ Integrate $\dfrac{dx}{(a^2 \cos^2x+b^2 \sin^2x)^2}$.
I can go up to the residue formula like in this example here
but then I just can't end up with the result for when $n=2$. I keep messing up my math. Can someone please show me how to get to the answer when $n=2$? I think I'v... | By setting $x=\arctan t$ we have:
$$I=\int\frac{dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}=\int\frac{1+t^2}{(a^2 t^2+b^2)^2}\,dt$$
and by setting $t=u\frac{b}{a}$ we get:
$$ I = \frac{1}{ab^3}\int\frac{1+\frac{b}{a} u^2}{(1+u^2)^2}\,du $$
so it is enough to check that:
$$\int \frac{du}{(1+u^2)^2} = \frac{1}{2}\left(\frac{u}{1+u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize $x^4+2x^2-x+2$? look at this:
$$x^4+2x^2-x+2$$
How to factorize it?
It should be changed to be in the form of standard factorization formulas.
| You can write
$$x^4+2x^2-x+2=(x^2+ax+b)(x^2+cx+d)$$
Then $a+c=0$, $bd=2$, $ad+bc=-1$ and $b+d+ac=2$. If you want integer values for $a,b,c,d$, you must have $b=\pm2$ or $\pm1$, and afters some checks,
$$x^4+2x^2-x+2=(x^2-x+1)(x^2+x+2)$$
Here is a more systematic approach. From the preceding equations, you have
$$c=-a$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all the complex numbers $z$ satisfying
Find all the complex numbers $z$ satisfying
$$
\bigg|\frac{1+z}{1-z}\bigg|=1
$$
So far I´ve done this:
$$
z=a+bi \\
\bigg|\frac{(1+a)+bi}{(1-a)-bi}\bigg|=1 \\
\mathrm{expression*conjugated} \\
\bigg|\frac{1+2bi-(a^2)-(b^2)}{1+2a+a^2+b^2}\bigg|=1
$$
| $\bf{My\; Solution}::$ Given $$\displaystyle \left|\frac{1+z}{1-z}\right| = 1\Rightarrow \left|1+z\right| = |1-z| = |z-1|.$$
Now Put $z=x+iy\;,$ Then $$|x+iy+1|=|x+iy-1|$$
So we get $$\sqrt{(x+1)^2+y^2} = \sqrt{(x-1)^2+y^2}\Rightarrow (x+1)^2=(x-1)^2$$.
So we get $x=0$. So all Complex no. are in the form of $z=iy\;,$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me conf... | There are many excellent proofs given above.
Another approach would be to note that the summation (LHS) is actually a binomial expansion of the result (RHS). (NB - This would only be applicable if you have to prove that the result is true, instead of having to find the result.)
Hence
$$\begin{align}
\color{blue}{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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No definite integrals of trigonometry I have big problems with the following integrals:
$$\int\frac{dx}{\sin^6 x+\cos^6x}$$
$$\int\frac{\sin^2x}{\sin x+2\cos x}dx$$
It isn't nice of me but I almost have no idea, yet I tried the trigonometric substitution $\;t=\tan\frac x2\;$ , but I obtained terrible things and can't d... | Let $$I = \int\frac{1}{\sin^6 x+\cos^6 x}dx = \int\frac{1}{(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x)}dx$$
So $$\int\frac{1}{\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x(\tan^2 x-1+\cot^2 x)}dx$$
So $$\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan^2 x+\cot^2 x-1)}dx$$
So $$I = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{... | Here's a way with trig (though it is harder than what you wrote).
Draw a right triangle with hypotenuse $\sqrt{2x}$. Put a side of length $1$ adjacent to the acute angle $\theta$. Then the opposite side is $\sqrt{2x-1}$. So $x=\sec(\theta)^2/2$,$dx=\sec(\theta)^2\tan(\theta) d \theta$, and $\sqrt{2x-1}=\tan(\theta)$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{... | Let $u = \sqrt{2} + \sqrt[3]{5}$, we have
$$\begin{align}
(u - \sqrt{2})^3 = 5
\iff & u^3 - 3\sqrt{2}u^2 + 6u - 2\sqrt{2} = 5\\
\implies &
\begin{cases}
\sqrt{2} &= \frac{u^3 + 6u - 5}{3u^2 + 2} \in \mathbb{Q}(u)\\
\sqrt[3]{5} &= u - \sqrt{2} \in \mathbb{Q}(u)
\end{cases}
\end{align}
$$
As a result, $\mathbb{Q}(\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the following series of transformations of inequalities holds? I am to calculate limit of the function $f(x,y)$ i am trying to apply squeeze theorem. Is the following series of transformations of this inequality correct? If not how to do this correctly? i.e. are those inequalities true or false? They should hold f... | Note that your last inequality is equivalent to the inequality
\begin{equation}
x + y \leq x^2 + y^2 \quad \forall x, y > 0,
\end{equation}
which is false (e.g. take $x, y = \frac{1}{2}$).
Perhaps use the following:
*
*For any $\alpha > 0$, $\ln(1 + \alpha) \leq \alpha$.
*$x^3 + y^3 \leq (x^2 + y^2)(x + y) \leq (x^... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
Linearly independent eigenvectors The matrix A=$\begin{pmatrix}0 & 1 &1 \\1& 0&1 \\ 1 & 1 & 0\end{pmatrix}$
has eigenvalues $\lambda=2$ with algebriac multiplicity $1$ and $\lambda=-1$ with multiplicity $2$
For $\lambda=-1$
$A+I=0 \implies$$\begin{pmatrix}1 & 1 &1 \\1& 1&1 \\ 1 & 1 & 1\end{pmatrix}\begin{pmatrix}x_1 \... | Compute
$$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does the Gaussian-Jordan elimination work when finding the inverse matrix? In order to find the inverse matrix $A^{-1}$, one can apply Gaussian-Jordan elimination to the augmented matrix
$$(A \mid I)$$
to obtain
$$(I \mid C),$$
where $C$ is indeed $A^{-1}$. However, I fail to see why this actually works, and readin... | Let me make it concrete in the following example(@pauly-b's answer is much better).
Suppose that 4 apples(any two are the same price) and 3 bananas(any two are the same price) in city A would cost 10 euros, and 3 apples and 2 bananas 7 euros. Let's calculate how much it would cost for 1 apple and 1 banana in A? We don'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Solve $x + y + z = xyz$ such that $x , y , z \neq0$ I came across the equation $x+y+z=xyz$ such that $x , y , z \neq 0$.
I set $x=1, y=2, z=3$ but how can i reach formal mathematical solution without " guessing " the answer ? Thank you
| Let $x = 1$, then $yz = y+z$.
Let $y = z$, then $y\cdot y = y^2 = 2y$.
$\frac{y^2}y = y = 2$.
Using substitution of $x=1$, $y=2$:
$1 + 2 + z = 2z$
$1 + 2 = 2z-z$
$z=3$
$1\times2\times3 = 1 + 2 + 3$
If $x = y = z$, $x$ to $z$ would equal $3^{\frac12}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Deriving the barycentric coordinates of a triangle's orthocenter, using the areal definition of such coordinates Wikipedia's "Altitude (triangle)" entry describes the barycentric coordinates of $\triangle ABC$'s orthocenter as $$(\tan A : \tan B : \tan C)$$
How would you prove this using solely the areal definition o... |
Barycentric Coordinates
In the triangle above,
$$
\begin{align}
CE&=\tan(A)\,AE\tag{1a}\\[6pt]
DE&=\cot(B)\,AE\tag{1b}\\
\frac{DE}{CE}&=\cot(A)\cot(B)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\triangle AEC$ has right angle at $E$
$\text{(1b)}$: $\triangle DEA\cong\triangle BGA$ and $\triangle DEA$ has a rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the closed-form for $\sum_{i=0}^n(-1)^i(\frac{1}{2})^i$ I start with simplifying:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2... | Here is the complete answer:
$$\sum_{i=0}^n(-1)^i(\frac{1}{2})^i=\sum_{i=0}^n(-\frac{1}{2})^i$$
then:
$$S = 1 + (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n$$
$$(-\frac{1}{2})S = (-\frac{1}{2}) + (-\frac{1}{2})^2 + ... +(-\frac{1}{2})^n+(-\frac{1}{2})^{n+1}$$
$$(-\frac{1}{2}-1)S = (-\frac{1}{2})^{n+1} - 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
| Since $1=\frac 33$, one has $$\begin{align}\frac{n(n+1)(n+2)}{3}+\color{blue}{1}\cdot (n+1)(n+2)&=\frac{n(n+1)(n+2)}{3}+\frac{\color{blue}{3}(n+1)(n+2)}{\color{blue}3}\\&=\frac{n\color{red}{(n+1)(n+2)}}{\color{red}{3}}+\frac{3\color{red}{(n+1)(n+2)}}{\color{red}{3}}\\&=\color{red}{\frac{(n+1)(n+2)}{3}}(n+3)\end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is upper Hessenberg form preserved through similarity transformation Suppose $X$ is non-singular and $M$ is upper Hessenberg.
Is $X^{-1}MX$ also upper Hessenberg.
| No.
Let
$$M
= \begin{pmatrix}
1 & 4 & 2 & 3 \\
3 & 4 & 1 & 7 \\
0 & 2 & 3 & 4 \\
0 & 0 & 1 & 3 \\
\end{pmatrix}
~~~~~~
X =
\begin{pmatrix}
1 & 0 & 0 & 1 \\
1 & 2 & 3 & -1 \\
1 & 2 & -1 & 0 \\
1 & 1 & 1 & 1
\end{pmatrix}
$$
then
$$X^{-1} =
\begin{pmatrix}
\frac{11}{7} & \frac{3}{7} & \frac{1}{7} & -\frac{8}{7} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Radius of a circle Let $f:\mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x)=x^2+1$. Find a real positive number $r$ such that the graph of $f$ intersects the circle about the origin and radius $r$ into two pieces , of which the one contains the origin and has an area equal to the value of the integral $ \displaystyle... | For first, by setting $x=t^2$:
$$ \int_{1}^{0}\frac{\log x}{\sqrt{x}} = 4\int_{0}^{1}(-\log t)\,dt = 4.\tag{1} $$
Next, let we assume that the circle and the parabola intersect in two points having ordinate $y>1$ and abscissa $x=\pm\sqrt{y-1}$. The area of the parabolic segment defined by these two points is:
$$ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.